The electron affinity of oxygen is \(-141 \mathrm{~kJ} / \mathrm{mol}\), corresponding to the reaction $$ \mathrm{O}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(g) $$ The lattice energy of \(\mathrm{K}_{2} \mathrm{O}(s)\) is \(2238 \mathrm{~kJ} / \mathrm{mol}\). Use these data along with data in Appendix \(\mathrm{C}\) and Figure 7.9 to calculate the "second electron affinity" of oxygen, corresponding to the reaction $$ \mathrm{O}^{-}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g) $$

To find the second electron affinity of oxygen, we must first understand the Born-Haber cycle and identify the components involved in the formation of potassium oxide (K2O). The Born-Haber cycle consists of the following components: 1. The formation of K and O into gaseous atoms from their standard states, ΔH₁ (sublimation of K and dissociation of O₂) 2. Ionization of gaseous potassium atoms to produce K⁺ ions, ΔH₂ (first ionization energy of K) 3. Electron affinity of gaseous oxygen atoms to produce O⁻ ions, ΔH₃ (first electron affinity of oxygen) 4. Second ionization of gaseous potassium atoms to produce K⁺ ions, ΔH₄ (second ionization energy of K) 5. Second electron affinity of gaseous O⁻ ions to produce O²⁻ ions, ΔH₅ (second electron affinity of oxygen) 6. The lattice energy formation of K2O, ΔH₆ (lattice energy of K2O)

Using Hess's Law, we can write the equation for the Born-Haber cycle as: ΔH = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + ΔH₅ + ΔH₆ We want to find ΔH₅, the second electron affinity of oxygen. Rearranging the equation: ΔH₅ = ΔH - (ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + ΔH₆)

We are given the following data: ΔH₃ (electron affinity of O) = -141 kJ/mol ΔH₆ (lattice energy of K₂O) = 2238 kJ/mol We also need data from Appendix C and Figure 7.9: ΔH₁ (sublimation of K) = 89 kJ/mol ΔH₂ (first ionization energy of K) = 419 kJ/mol ΔH₄ (second ionization energy of K) = 3052 kJ/mol Finally, we need the standard enthalpy of formation of K2O (ΔH): ΔH (Enthalpy of formation of K₂O) = -363.8 kJ/mol

Now we can substitute the values into the equation and calculate the second electron affinity of oxygen: ΔH₅ = (-363.8 kJ/mol) - (89 kJ/mol + 419 kJ/mol - 141 kJ/mol + 3052 kJ/mol + 2238 kJ/mol) ΔH₅ = -363.8 - 4657 ΔH₅ = 4293.2 kJ/mol The second electron affinity of oxygen is approximately 4293.2 kJ/mol.